I've taken a course Complex Analysis, but I don't understand why the phrase "For Uniform Convergence on Compact Sets" was used all the time. I always got the impression this was "good enough" for something. But for what? Why is it enough? Why do we care? When is it enough? When is it not enough?

Did this same idea pop up somewhere in real analysis/measure theory?

I really appreciate your exposition on this matter!

  • 4
    $\begingroup$ It is weaker than "uniform convergence" and stronger than pointwise convergence, and if a sequence of holomorphic functions converges "uniformly on compact sets", then the limit is also holomorphic (by Morera's theorem). This is the simplest reason I can think of. $\endgroup$ – Prahlad Vaidyanathan May 14 '14 at 16:21

In essence, the key idea is that given an open region $D$, a sequence of functions $\{f_n\}$ analytic on $D$, and a function $f$ such that $\{f_n\}$ converges to $f$ uniformly on every compact subset of $D$, then the function $f$ is itself analytic.

Sometimes we say "$\{f_n\} \to f$ on compacta," though I hate this phrasing.

Ok, that's a lot of hypotheses. Why does it matter?

Convergence on compacta is necessary to set up some key results in complex analysis, for instance, Hurwitz' theorem.

In the end, uniform convergence on compacta shows that analyticity is preserved under uniform limits, which underscores the idea that a function $f:\mathbb{C} \to \mathbb{C}$ that is differentiable once is differentiable infinitely many times. This is one of the key differences between analysis on $\mathbb{C}$ and $\mathbb{R}^2$.

  • $\begingroup$ Can you explain what you mean by "analyticity is preserved under uniform limits?" $\endgroup$ – MathStudent May 14 '14 at 16:31
  • 1
    $\begingroup$ Sure, it's very simple. Since $f = \lim_n f_n$, then $\int_\Gamma f = \int_\Gamma \lim_n f_n = \lim_n \int_\Gamma f_n$. Now, of course since each $f_n$ is analytic, then we must have $\int_\Gamma f_n = 0$, so $\int_\Gamma f = 0$ and $f$ is analytic. $\endgroup$ – Emily May 14 '14 at 16:34
  • $\begingroup$ In other words, taking the uniform limit of the analytic sequence $\{f_n\}$ doesn't ever annihilate analyticity. $\endgroup$ – Emily May 14 '14 at 16:35
  • $\begingroup$ In the reals, however, we have have a sequence of differentiable functions $\{g_n\} \to g$ where $g$ is not everywhere differentiable (in fact, it can be nowhere differentiable). $\endgroup$ – Emily May 14 '14 at 16:36
  • $\begingroup$ OK. What exactly is meant by "taking the uniform limit" as opposed to "taking a limit." And what part of all this is allowing you to pass the limit inside the integral? $\endgroup$ – MathStudent May 14 '14 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.